Răspuns:
1.S=2/5+1/5+1/10+1/20+...+1/5·2ⁿ⁻⁷=
2/5+1/5(1+1/2+1/2²+...+1/2ⁿ⁻⁷)
Notam s=1+1/2+1/2²+...+1/2ⁿ⁻⁷
inmultesti egalitatea cu 2
2s=2+1+1/2+1/2²+...+1/2ⁿ⁻⁷+1/2ⁿ⁻⁸
Calculebzi
2s-s=2+1+1/2+1/2²+...+1/2ⁿ⁻⁷+1/2ⁿ⁻⁸-1-1/2-1/2²-....-1/2ⁿ⁻⁷=
s=2-0-0...-1/2ⁿ⁻⁸
s=2-1/2ⁿ⁻⁸
S=2/5+1/5(2-1/2ⁿ⁻⁸)
S=2/5+2/5-1/2ⁿ⁻⁸·5
S=4/5-1/2ⁿ⁻⁸·5
S=1/5(4-1/2ⁿ⁻⁸)
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2.(x1,x2,x3,...,x2015}={1,2,3,...,2015}
Observi ca in ambele paranteze sunt cate 2015 elemente
lx1-1l,lx2-2l,...,lx2015-2015l sunt 2015 numere
Dar aceste diferente nu pot depasi 2014 deci
lxi-2015l≤2014 i∈{1,2,...,2015}
deci avem 2015 numere care aiau valori de la 1 la 2014 deciexista 2 numere care se repeta
Explicație pas cu pas:
